Find the surface area of a surface created by rotating the region bounded by 𝑓(𝑥) = 𝑥2 and the x-axis, on [0,1], about the x-axis?

To find the surface area of a surface created by rotating a curve around the x-axis, we can use the formula for the surface area of revolution:

A = 2π ∫ from a to b [f(x) * sqrt(1 + (f'(x))^2)] dx

Here, f(x) = x^2 is the function that defines the curve, and f'(x) = 2x is its derivative. The limits of integration a and b are 0 and 1, respectively.

Let's plug these into the formula:

A = 2π ∫ from 0 to 1 [x^2 * sqrt(1 + (2x)^2)] dx

This integral is a bit tricky to solve, but we can simplify the integrand by factoring out x^2:

A = 2π ∫ from 0 to 1 [x^2 * sqrt(1 + 4x^2)] dx

Now, we can use a substitution to solve the integral. Let u = 1 + 4x^2. Then, du = 8x dx, and dx = du / (8x).

Substituting these into the integral gives:

A = 2π ∫ from 1 to 5 [(1/8) * sqrt(u)] du

This integral can be solved using the power rule for integration, which gives:

A = 2π * [(2/3) * (5^(3/2) - 1^(3/2))]

Solving this gives:

A = 2π * [(2/3) * (5sqrt(5) - 1)]

So, the surface area of the surface created by rotating the region bounded by f(x) = x^2 and the x-axis, on [0,1], about the x-axis is 2π * [(2/3) * (5sqrt(5) - 1)].